Leg and altitude as a fraction of the hypotenuse

Text

In triangle $A\hat{C}B$, right-angled at $\hat{C}$, the leg $\overline{AC}$ equals $3$, while the altitude to the hypotenuse is one quarter of the hypotenuse. Determine all the sides of the triangle.

Solution

Let $\overline{BC}=b$ and $\overline{AB}=c=\sqrt{9+b^2}$. The altitude to the hypotenuse is $\overline{CH}=\dfrac{\overline{AC}\cdot\overline{BC}}{\overline{AB}}=\dfrac{3b}{c}$. The condition $\overline{CH}=\dfrac14 c$ gives $\frac{3b}{c}=\frac{c}{4}\ \Longrightarrow\ 12b=c^2=9+b^2\ \Longrightarrow\ b^2-12b+9=0.$ Solving: $b=\frac{12\pm\sqrt{108}}{2}=6\pm3\sqrt3.$ Both roots are positive: the problem has two solutions.

• $\overline{BC}=6-3\sqrt3\approx0{,}80$, whence $\overline{AB}=6\sqrt{2-\sqrt3}\approx3{,}11$;
• $\overline{BC}=6+3\sqrt3\approx11{,}20$, whence $\overline{AB}=6\sqrt{2+\sqrt3}\approx11{,}59$.

$\boxed{\overline{BC}=6\pm3\sqrt3,\quad \overline{AB}=6\sqrt{2\pm\sqrt3}\ \ (\overline{AC}=3).}$